Question: Find the distance between the vertices of the hyperbola $9x^2 + 54x - y^2 + 10y + 55 = 0.$
Solution: Completing the square in $x$ and $y,$ we get
\[9(x + 3)^2 - (y - 5)^2 = 1,\]which we can write as
\[\frac{(x + 3)^2}{1/9} - \frac{(y - 5)^2}{1} = 1.\]Thus, $a^2 = \frac{1}{9}$ and $b^2 = 1,$ so $a = \frac{1}{3}.$  Therefore, the distance between the vertices is $2a = \boxed{\frac{2}{3}}.$